Calculus of Trigonometry Functions

\(\cfrac{d}{dx} \sin \left[ f(x) \right] = f'(x) \cos \left[ f(x) \right] \)

\(\cfrac{d}{dx} \cos \left[ f(x) \right] = -f'(x) \sin \left[ f(x) \right] \)

\(\int f'(x) \sin \left[ f(x) \right] \ dx = -\cos \left[ f(x) \right] +c \)

\(\int f'(x) \cos \left[ f(x) \right] \ dx = \sin \left[ f(x) \right] +c \)

Deriving \(y = \sin(x)\) using first principles:

\(\begin{aligned} \frac{dy}{dx} &= \lim_{h \rightarrow 0} \left[ \frac{f(x+h)-f(x)}{h} \right] \\[12pt] &= \lim_{h \rightarrow 0} \left[ \frac{\sin(x+h)-\sin(x)}{h} \right] \\[12pt] &= \lim_{h \rightarrow 0} \left[ \frac{\sin x\cos h+\cos x\sin h-\sin x}{h} \right] \\[12pt] &= \lim_{h \rightarrow 0} \left[ \frac{\sin x (\cos h-1)+\cos x\sin h}{h} \right] \\[12pt] &= \lim_{h \rightarrow 0} \left[ \frac{\sin x (\cos h-1)}{h}+ \frac{\cos x\sin h}{h} \right] \\[12pt] &= \lim_{h \rightarrow 0} \left[ \sin x \left[ \frac{ (\cos h-1)}{h} \right] + \cos x \left[ \frac{\sin h}{h} \right] \right] \\[12pt] &= \cos x \end{aligned} \)

Deriving \(y = \cos(x)\) using first principles:

\(\begin{aligned} \frac{dy}{dx} &= \lim_{h \rightarrow 0} \left[ \frac{f(x+h)-f(x)}{h} \right] \\[12pt] &= \lim_{h \rightarrow 0} \left[ \frac{\cos (x+h)-\cos x}{h} \right] \\[12pt] &= \lim_{h \rightarrow 0} \left[ \frac{(\cos x\cos h - \sin x\sin h)-\cos x}{h} \right] \\[12pt] &= \lim_{h \rightarrow 0} \left[ \frac{\cos x(\cos h - 1)-\sin x\sin h}{h} \right] \\[12pt] &= \lim_{h \rightarrow 0} \left[ \frac{\cos x(\cos h - 1)}{h} - \frac{\sin x\sin h}{h} \right] \\[12pt] &= \lim_{h \rightarrow 0} \left[ \cos x\left[\frac{(\cos h - 1)}{h} \right] - \sin x \left[ \frac{\sin h}{h} \right] \right] \\[12pt] &= \cos x\left(0 \right) - \sin x \left( 1 \right) \\[12pt] &= -\sin x \end{aligned} \)

1) Calculate \(\cfrac{d}{dx} (\cos 3x + 2x) \)

\( \ \ \ \ \ \ = -3\sin 3x + 2\)

2) Calculate \(\cfrac{d}{dx} (\cos \frac{x}{2} - 2e^{\sin (x^2+3)})\)

\( \ \ \ \ \ \ \begin{aligned} &= -\frac{1}{2} \sin \frac{x}{2} - \frac{d}{dx} (2e^{\sin(x^2+3)}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(x) = \sin (x^2+3) \\[10pt] &= - \frac{1}{2} \sin \frac{x}{2} - 2(2x) \cos (x^2+3) e^{\sin (x^2+3)} \ \ \ \ f'(x) = 2x \cos (x^2+3) \\[10pt] &= - \frac{1}{2} \sin \frac{x}{2} - 4x\cos (x^2+3) e^{\sin (x^2+3)} \end{aligned} \)

3) Calculate \(\displaystyle \int_{0}^{\frac{\pi}{2}} (\sin 2x) (\cos 2x)^4 \ dx\)

\(\displaystyle \begin{aligned} &= - \frac{1}{2} \int_{0}^{\frac{\pi}{2}} -2(\sin 2x)(\cos 2x)^4 \ dx \\[10pt] &= - \frac{1}{2} \times \left[\cfrac{(\cos 2x)^5}{5} \right]_{0}^{\frac{\pi}{2}} \\[10pt] &= - \frac{1}{2} \times \left[(\frac{(\cos \pi) ^ 5}{5})-(\frac{(\cos 0)^5}{5}) \right] \\[10pt] &= -\frac{1}{2} \times \left[ \frac{(-1)^5}{5} - \frac{1}{5} \right] \\[10pt] &= - \frac{1}{2}\left[ -\frac{2}{5} \right] \\[10pt] &= \frac{1}{5} \end{aligned} \)